Problem 28 - Precalculus

Question

In triangle \(O A B,\) lengths \(O A=O B=6\) in. and \(\angle A O B=72^{\circ} .\) Find \(A B .\) Ilint: Draw a perpendicular from \(O\) to \(A B\). Round the answer to one decimal place.

Step-by-Step Solution

Answer

The length of AB is approximately 9.7 inches.

Step 1: Understand the Given Triangle

We start with triangle \( OAB \), where \( OA = OB = 6 \) inches, and \( \angle AOB = 72^\circ \). This implies that \( O \) is the center of a circle, making triangle \( OAB \) isosceles with sides \( OA \), \( OB \) equal.

Step 2: Draw the Perpendicular from O to AB

As suggested, draw a line from \( O \), perpendicular to \( AB \), meeting \( AB \) at a point \( P \). This creates two right triangles, \( OAP \) and \( OBP \), and splits the angle \( \angle AOB \) into two equal \( 36^\circ \) angles since a perpendicular from the center bisects the angle.

Step 3: Calculate the Half-Length of AB

In right triangle \( OAP \), the cosine of \( \angle AOP \) (which is \( 36^\circ \)) can be used to find \( AP \). The cosine formula is \( \cos(36^\circ) = \frac{AP}{OA} \). Substitute \( OA = 6 \) to get \( AP = 6\cos(36^\circ) \).

Step 4: Double the Half-Length for AB

Since \( AP = PB \) (because \( P \) is the midpoint of \( AB \)), \( AB = 2 \times AP = 12\cos(36^\circ) \).

Step 5: Calculate Cosine and Finalize Length

Find \( \cos(36^\circ) \), which approximately equals 0.8090. Therefore, \( AB = 12 \times 0.8090 = 9.708 \) inches. Rounding this to one decimal place gives \( AB \approx 9.7 \) inches.

Key Concepts

Isosceles Triangle

An isosceles triangle is a special type of triangle with two sides of equal length. In triangle geometry, these two equal sides make the base angles (angles opposite the equal sides) also equal. In our problem, triangle \( OAB \) is an isosceles triangle because \( OA = OB = 6 \) inches. This equality helps in simplifying calculations, especially when solving for unknown sides using trigonometric rules.

• Identifying an isosceles triangle is crucial, as it often allows the use of symmetry to simplify problems.
• In this exercise, the isosceles triangle helps in discovering that the perpendicular from \( O \) to \( AB \) bisects the angle \( \angle AOB \) into two equal parts \( 36^\circ \).

Breaking down isosceles triangles can be quite useful in geometric problems, providing opportunities to apply specific rules and properties that might not otherwise be evident.

Cosine Rule

The cosine rule, also known as the law of cosines, is an essential tool in trigonometry that relates the lengths of the sides of a triangle to the cosine of one of its angles. It is a generalization of the Pythagorean theorem used in any kind of triangle. The formula is given by:\[c^2 = a^2 + b^2 - 2ab \cos(C)\\]where \( a \), \( b \), and \( c \) are the sides of a triangle, and \( C \) is the angle opposite side \( c \).
In the problem given:

• We used trigonometric properties, rather than the direct Cosine Rule, to calculate the length of \( AB \).
• You can use the cosine function definition: \( \cos(\theta) = \frac{adjacent}{hypotenuse} \).

With the angle \( 36^\circ \) and side \( OA = 6 \), it is direct to calculate \( AP \), which is half of \( AB \). Hence, the total length of \( AB \) was calculated using the cosine of \( 36^\circ \). This approach simplifies the problem without delving deep into full-blown cosine rule application for scalene triangles.

Right Triangle

A right triangle is a triangle in which one of the angles is exactly 90 degrees. In this exercise, drawing a perpendicular line from \( O \) to \( AB \) divides \( \triangle OAB \) into two right triangles: \( \triangle OAP \) and \( \triangle OBP \).

• Right triangles have unique properties, such as the Pythagorean theorem and the basic trigonometric ratios: sine, cosine, and tangent.
• In right triangle \( OAP \), the hypotenuse is \( OA \), and you can use trigonometric functions like cosine to find adjacent and opposite sides.

By solving right triangle \( OAP \) with trigonometric ratios, the problem asks us to focus on the cosine of \( 36^\circ \), allowing us to directly calculate \( AP \). Once \( AP \) is calculated, doubling it provides the whole length of \( AB \), due to symmetry. These steps clearly illustrate how understanding right triangles can significantly aid in resolving geometric problems based upon trigonometric principles.